最小割 D. Behind the Wall Samara University ACM ICPC 2016-2017 Quarterfinal Qualification Contest

题目链接：http://codeforces.com/gym/101149/problem/D

题目大意：

堡垒受到攻击。堡垒是n*m的矩阵，矩阵里刚开始都是平地，然后那个数值表示在当前平地上建一面墙需要a[i][j]的时间。目前我们在位置(r, c)，我们找一种方法，把(r,c)全部围起来需要的最短时间？

思路：拆点，拆成in和out两个，in和out之间的cap就是a[i][j]，然后就是简单的建边拉。

//看看会不会爆int!数组会不会少了一维！
//取物问题一定要小心先手胜利的条件
#include <bits/stdc++.h>
using namespace std;
#pragma comment(linker,"/STACK:102400000,102400000")
#define LL long long
#define ALL(a) a.begin(), a.end()
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define haha printf("haha
")

const int maxn = 50 * 50 * 2 + 5;
const int INF = 0x3f3f3f3f;
struct Edge {
  int from, to, cap, flow;
};

struct Dinic {
  int n, m, s, t;        ///节点的个数，边的编号，起点，终点
  vector<Edge> edges;    // 边数的两倍
  vector<int> G[maxn];   // 邻接表，G[i][j]表示结点i的第j条边在e数组中的序号
  bool vis[maxn];        // BFS使用
  int d[maxn];           // 从起点到i的距离
  int cur[maxn];         // 当前弧指针
  /////////蓝书363
  int inq[maxn];         // 是否在队列中
  int p[maxn];           // 上一条弧
  int a[maxn];           //可改进量

  void ClearAll(int n) {
    this->n = n;        ///这个赋值千万不能忘
    for(int i = 0; i < n; i++) G[i].clear();
    edges.clear();
  }

  void ClearFlow() {  ///清除流量,例如蓝书368的UVA11248里面的优化,就有通过清除流量来减少增广次数的
    for(int i = 0; i < edges.size(); i++) edges[i].flow = 0;
  }

  void Reduce() {///直接减少cap,也是减少增广次数的
    for(int i = 0; i < edges.size(); i++) edges[i].cap -= edges[i].flow;
  }

  void AddEdge(int from, int to, int cap) {
    edges.push_back((Edge){from, to, cap, 0});
    edges.push_back((Edge){to, from, 0, 0});
    m = edges.size();
    G[from].push_back(m-2);
    G[to].push_back(m-1);
  }

  bool BFS() {///bfs构建层次图
    memset(vis, 0, sizeof(vis));
    queue<int> Q;
    Q.push(s);
    vis[s] = 1;
    d[s] = 0;
    while(!Q.empty()) {
      int x = Q.front(); Q.pop();
      for(int i = 0; i < G[x].size(); i++) {
        Edge& e = edges[G[x][i]];
        if(!vis[e.to] && e.cap > e.flow) {//只考虑残量网络中的弧
          vis[e.to] = 1;
          d[e.to] = d[x] + 1;
          Q.push(e.to);
        }
      }
    }
    return vis[t];
  }

  int DFS(int x, int a) {///a表示目前为止，所有弧的最小残量。但是也可以用它来限制最大流量，例如蓝书368LA2957，利用a来保证增量，使得最后maxflow=k
    if(x == t || a == 0) return a;
    int flow = 0, f;
    for(int& i = cur[x]; i < G[x].size(); i++) {//从上次考虑的弧，即已经访问过的就不需要在访问了
      Edge& e = edges[G[x][i]];
      if(d[x] + 1 == d[e.to] && (f = DFS(e.to, min(a, e.cap-e.flow))) > 0) {
        e.flow += f;
        edges[G[x][i]^1].flow -= f;
        flow += f;
        a -= f;
        if(a == 0) break;//如果不在这里终止，效率会大打折扣
      }
    }
    return flow;
  }
    /**最大流*/
  int Maxflow(int s, int t) {
    this->s = s; this->t = t;
    int flow = 0;
    while(BFS()) {
      memset(cur, 0, sizeof(cur));
      flow += DFS(s, INF);///这里的INF可以发生改变，因为INF保证的是最大残量。但是也可以通过控制残量来控制最大流
    }
    return flow;
  }

    /**最小割*/
  char ch[maxn][maxn];
  void Mincut(int x, int y) { /// call this after maxflow求最小割，就是S和T中还存在流量的东西
    BFS();///重新bfs一次
    for (int i = 0; i < x; i++)
        for (int j = 0; j < y; j++)
            ch[i][j] = '.';
    int cnt = 0;
    for(int i = 0; i < edges.size(); i++) {
      Edge& e = edges[i];
      if(vis[e.from] && !vis[e.to] && e.cap >= 0 && e.to - e.from == x * y) {///这里和ISAP不一样
          cnt++;
          int nx = e.from / y, ny = e.from - nx * y;
          ch[nx][ny] = 'X';
          //printf("e.from = %d e.to = %d nx = %d ny = %d
", e.from, e.to, nx, ny);
      }
    }
    //printf("cnt = %d
", cnt);
    for (int i = 0; i < x; i++){
        for (int j = 0; j < y; j++){
            printf("%c", ch[i][j]);
        }
        cout << endl;
    }

  }

  void debug(){///debug
    for (int i = 0; i < edges.size(); i++){
        printf("u = %d v = %d cap = %d flow = %d
", edges[i].from + 1, edges[i].to + 1, edges[i].cap, edges[i].flow);
    }
  }
};
Dinic g;

int n, m, a, b;
int atlas[maxn][maxn];

int dx[] = {1, -1, 0, 0};
int dy[] = {0, 0, 1, -1};
int in_id(int x, int y){
    return x * m + y;
}
int out_id(int x, int y){
    return n * m + x * m + y;
}

int main(){
    scanf("%d%d%d%d", &n, &m, &a, &b);
    a--, b--;
    for (int i = 0; i < n; i++){
        for (int j = 0; j < m; j++){
            scanf("%d", &atlas[i][j]);
        }
    }
    int s = in_id(a, b), t = n * m * 2 + 1;
    g.ClearAll(t);
    for (int i = 0; i < n; i++){
        for (int j = 0; j < m; j++){
            if (a == i && b == j) {
                for (int k = 0; k < 4; k++){
                    int nx = i + dx[k], ny = j + dy[k];
                    if (nx < 0 || ny < 0 || nx >= n || ny >= m) continue;
                    g.AddEdge(in_id(i, j), in_id(nx, ny), INF);
                }
                continue;
            }
            else {
                g.AddEdge(in_id(i, j), out_id(i, j), atlas[i][j]);
                if (i == 0 || j == 0 || i == n-1 || j == m-1)
                    g.AddEdge(out_id(i, j), t, INF);
            }
            for (int k = 0; k < 4; k++){
                int nx = i + dx[k], ny = j + dy[k];
                if (nx < 0 || ny < 0 || nx >= n || ny >= m) continue;
                if (nx == a && ny == b) continue;
                g.AddEdge(out_id(i, j), in_id(nx, ny), INF);
            }
        }
    }
    printf("%d
", g.Maxflow(s, t));
    g.Mincut(n, m);
    return 0;
}